Question

In a Young's double slit experiment, the ratio of the slit's width is 4:1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be
A. 65:1
B. 9:1
C. 4:1
D. 25:1

Answer 1

Hint: Here, we use the formula of maximum and minimum intensities and applying the given values will give the resultant. At last, the ratio of maximum intensity to minimum intensity is solved, which gives the required answer.

Formula used:
$\eqalign{
  & {I_{max}} = {(\sqrt {{I_{1}}} + \sqrt {{I_2}} )^2} \cr
  & {I_{min}} = {(\sqrt {{I_{1}}} - \sqrt {{I_{2}}} )^2} \cr} $

Complete step by step answer:
We know that the square of amplitude of a wave gives us intensity. The maximums and minimum intensity can be calculated by using the formula.
$\eqalign{
  & {I_{1}} = 4{I_{0}} \cr
  & {I_2} = {I_0} \cr
  & {I_{max}} = {(\sqrt {{I_{1}}} + \sqrt {{I_2}} )^2} \cr
  & = {(2\sqrt {{I_0}} + \sqrt {{I_{0}}} )^2} = 9{I_0} \cr
  & {I_{min}} = {(\sqrt {{I_{1}}} - \sqrt {{I_{2}}} )^2} \cr
  & = (2\sqrt {{I_0}} - \sqrt {{I_{0}}} )2 = {I_{0}} \cr
  & \therefore \dfrac{{{I_{max}}}}{{{I_{min}}}} = \dfrac{9}{1} \cr} $

So, the correct answer is “Option B”.

Additional Information: As we know, Young’s double-slit experiment uses two coherent sources of light placed at a small distance apart. These two sources of light are a few orders of magnitude greater than the wavelength of light is used. Young’s double-slit experiment helps us in understanding the superposition of light waves, also it tells about the type of interference taking place. Constructive and destructive are the two types of inference, that can be observed at the screen which is placed at a few distance. A screen or photo detector is placed at a large distance ’D’ away from the slits.
Due to the distance between the two light sources, there is some path difference observed. This is the path difference between two meetings at a point on the screen. Due to this path difference in Young’s double slit experiment, some points on the screen are bright and some points are dark.
Therefore, we finally observe the constructive and destructive fringes on the screen.
Further, intensity is defined as the power radiated per unit area. Here the area is measured on the plane perpendicular to the direction of propagation of the energy. The SI unit of intensity is measured in watts per square meter.

Note: When we calculate the ratio of any physical quantity, there is no resultant unit. In Young’s-Double slit experiment the light source is coherent. Lasers can be used as the light source because of its coherence property. The intensity can be calculated as the square of the amplitude.